Suppose $X$ and $Y$ are two (smooth, affine) algebraic varieties. Let $\mathcal{F}$ be a locally free coherent sheaf over $X \times Y$, and let $\mathcal{G}$ be the pushforward of $\mathcal{F}$ to $X$. Is it true that $\mathcal{G}$ is a locally free quasicoherent sheaf?
[Math] Does a locally free sheaf over a product pushforward to a locally free sheaf
ac.commutative-algebraag.algebraic-geometry
Best Answer
The answer is "yes" (though I can't imagine a situation where one would really need this fact). More generally, if $A$ and $B$ are arbitrary commutative algebras over a field $k$ with $A$ noetherian and if $M$ is an $A \otimes_k B$-module which is locally free as such (perhaps not finitely generated) then $M$ is locally free as an $A$-module. Here, by "locally free" I meant relative to the Zariski topology.
Without loss of generality (since $A$ is noetherian), we may and do assume that ${\rm{Spec}}(A)$ is connected.
The first thing to observe is that $M$ is projective as an $A \otimes_k B$-module. Indeed, projectivity is a Zariski-local (even fpqc-local) property for modules over commutative rings, by 3.1.3 part II of Raynaud-Gruson (and the fact that faithfully flat ring maps satisfy their condition (C), using 3.1.4 part I of Raynaud-Gruson), so any locally free module over a commutative ring is projective. Thus, $M$ is a direct summand of a free $A \otimes_k B$-module, which in turn is also free as an $A$-module. Hence, $M$ is projective as a $A$-module. If $M$ is module-finite as such then it is certainly locally free (since $A$ is noetherian). But if it is not module-finite then we're again done since $A$ is noetherian with connected spectrum, as then it follows that any projective $A$-module that is not finitely generated is free! This is Bass' theorem "big projective modules are free"; see Corollary 4.5 in his paper with that title.